First-order linear differential equation

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Definition

Format of the differential equation

A first-order linear differential equation is a differential equation of the form:

\frac{dy}{dx} + p(x)y = q(x)

where p,q are known functions.

Solution method and formula: indefinite integral version

Let H(x) be an antiderivative for p(x), so that H'(x) = p(x). Then, we multiply both sides by e^{H(x)}. Simplifying, we get:

\frac{d}{dx}[e^{H(x)}y] = q(x)e^{H(x)}

Integrating, we get:

e^{H(x)}y = \int q(x)e^{H(x)} \, dx

Rearranging, we get:

y = e^{-H(x)}\int q(x)e^{H(x)} \, dx where H is an antiderivative of p.

In particular, we obtain that:

\mbox{General solution} = \mbox{Particular solution} + Ce^{-H(x)}, C \in \R

The function e^{H(x)} is termed the integrating factor for the differential equation because multiplying by this turns the differential equation into an exact differential equation, i.e., a differential equation to which we can apply integration on both sides.

Solution method and formula: definite integral version

Suppose we are given the initial value condition that at x = x_0, y = y_0.

Let H(x) be an antiderivative for p(x), so that H'(x) = p(x). Then, we multiply both sides by e^{H(x)}. Simplifying, we get:

\frac{d}{dx}[e^{H(x)}y] = q(x)e^{H(x)}

Integrating from x_0 to (arbitrary) x, we get:

e^{H(x)}y - e^{H(x_0)}y_0= \int_{x_0}^x q(t)e^{H(t)} \, dt

Thus, the general expression is:

y = e^{-H(x)}\left(e^{H(x_0)}y_0 + \int_{x_0}^x q(t)e^{H(t)} \, dt\right)

Examples

Simple example

Consider the differential equation:

y' + y = e^{e^x}

Here, p(x) = 1, q(x) = e^{e^x}. Take H(x) =x and get:

y = e^{-x} \int e^{e^x}e^x \, dx

This gives:

y = e^{-x}(e^{e^x} + C), \qquad C \in \R

Example that is better solved by subtitution

Consider:

xy' + y = \sin x

Divide both sides by x to get:

y' + \frac{y}{x} = \frac{\sin x}{x}

This is linear, with p(x) = 1/x, q(x) = (\sin x)/x. Take H(x) = \ln x and e^{H(x)} = x (see note):

y = \frac{1}{x} \int \frac{\sin x}{x} x \, dx

This gives:

y = \frac{C - \cos x}{x}, \qquad C \in \R

The linear method is unnecessary -- we divided and multiplied by x. A better solution would be to substitute u = xy and get a separable differential equation.

Example where a particular solution is obtained by inspection

Consider:

y' + y = \tan x + \tan^2x

The linear method gives:

y = e^{-x} \int e^x(\tan x + \tan^2x) \, dx

The integration is not easy. So, instead of trying to do the integration directly, we note that the answer is:

y = \mbox{Particular solution} + Ce^{-x}

It thus suffices to find a particular solution. Inspection and guesswork gives a solution y = \tan x - 1. The general solution is thus:

y = \tan x - 1 + Ce^{-x}